Algebraic Differentials Separability and Smooth Local Presentations — Examples

The quotient formula. With $f=y^2-x^3$ one has $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-3x^2$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=2y$ on the monomial basis [F1], so the Jacobian matrix of the single equation is the row $(-3x^2,2y)$ and ${\mathrm{\Omega }}_{A\mathrm{/}k}\cong A^2\mathrm{/}A\cdot (-3x^2,2y)=(A\mathrm{d}x\oplus A\mathrm{d}y)\mathrm{/}(2y\mathrm{d}y-3x^2\mathrm{d}x)$, exactly as displayed; no injectivity of the conormal map is used or claimed [F1].

The curve has dimension one. The subring $k[x]\subseteq A$ is a polynomial ring, and $A$ is a finite $k[x]$-module with basis $1,y$ because $y^2=x^3\in k[x]$, so the extension is integral and injective and $A\mathrm{\ne }0$; hence $\dim A=\dim k[x]=1$ by [F2] and [F3].

The fibre at the origin. The maximal ideal $\mathfrak{m}=(x,y)A$ corresponds to the origin, with residue field $\kappa (\mathfrak{m})=k$ because $A\mathrm{/}\mathfrak{m}=k$; tensoring the presentation of step 1.1 with $A\to k$ gives ${\mathrm{\Omega }}_{A\mathrm{/}k}{\otimes }_{A}k\cong k^2\mathrm{/}k\cdot (0,0)=k^2$, as both coefficients $2y$ and $3x^2$ vanish at the origin even when $2=0$ or $3=0$ in $k$. Thus the cotangent fibre at the origin has dimension two, equal to the number of variables, while $A$ has dimension one by step 1.2.

The source. By [F2] the module ${\mathrm{\Omega }}_{k[x]\mathrm{/}k}=k[x]\cdot \mathrm{d}x$ is free of rank one, so $k{\otimes }_{k[x]}{\mathrm{\Omega }}_{k[x]\mathrm{/}k}\cong k{\otimes }_{k[x]}k[x]\cong k$ by [F4], a one-dimensional $k$-vector space with basis $1\otimes \mathrm{d}x$; in particular $1\otimes \mathrm{d}x\mathrm{\ne }0$.

The target and the map. The map $\epsilon$ exhibits $k$ as a $k$-algebra over itself, so every element of $k$ lies in the image of the base ring and [F1] gives ${\mathrm{\Omega }}_{k\mathrm{/}k}=0$. The canonical map of [F3] sends $1\otimes \mathrm{d}x$ to $\mathrm{d}\epsilon (x)=\mathrm{d}0=0$, hence is the zero map from a one-dimensional space to the zero space: not injective, and not an isomorphism. The hypothesis of a general algebra map in the base-change statement is therefore essential.

The separable case. By [F1] write $E=k(\alpha )$, and let $P\in k[X]$ be the monic minimal polynomial of $\alpha$, so that $E\cong k[X]\mathrm{/}(P)$ by [F2]. Since $E\mathrm{/}k$ is separable the element $\alpha$ is separable, so $P$ is separable and $\gcd (P,P^{\mathrm{\prime }})=1$ by [F4]; as $P(\alpha )=0$ and $P^{\mathrm{\prime }}(\alpha )=0$ would force $P\mid P^{\mathrm{\prime }}$ by [F2], impossible for $0\mathrm{\ne }{P}^{\mathrm{\prime }}$ of degree less than $\mathrm{deg}P$, we get $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$. By [F3] applied to the presentation $E=k[X]\mathrm{/}(P)$ the module ${\mathrm{\Omega }}_{E\mathrm{/}k}$ is $E\mathrm{/}(P^{\mathrm{\prime }}(\alpha ))$, and $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ is a unit of the field $E$, so ${\mathrm{\Omega }}_{E\mathrm{/}k}=0$.

The polynomial $t^p-a$ is irreducible. Let $Q$ be a monic irreducible factor of $t^p-a$ [F5] and put $F=k[t]\mathrm{/}(Q)$, a field with class $x$ of $t$ satisfying $x^p=a$ [F5]. By [F6] the identity $(t-x{)}^p=t^p-x^p=t^p-a$ holds in $F[t]$, so every root of $Q$ in a splitting field is a root of $(t-x{)}^p$, hence equals $x$: the polynomial $Q$ has exactly one distinct root. If $\mathrm{deg}Q=1$ then $x\in k$ and $a=x^p\in k^p$, contrary to the hypothesis, so $\mathrm{deg}Q\ge 2$; and $\mathrm{deg}Q\ge 2$ with $Q^{\mathrm{\prime }}\mathrm{\ne }0$ would make $Q$ separable with $\mathrm{deg}Q$ distinct roots by [F4], a contradiction. Hence $Q^{\mathrm{\prime }}=0$, which in characteristic $p$ means that $Q$ is a polynomial in $t^p$ and, since $0\mathrm{\ne }Q$ has degree at most $p$, forces $Q=t^p-a$; so $t^p-a$ is irreducible, $L=k[t]\mathrm{/}(t^p-a)$ is a field by [F5], and $t\in L$ satisfies $t^p=a\mathrm{\notin }{k}^p$ hence $t\mathrm{\notin }k$.

The differentials of the inseparable field. By [F3] applied to the single equation $t^p-a$ with $A=k$ and $B=L$, and by the derivative computation $P^{\mathrm{\prime }}(t)=p{t}^{p-1}=0$ of [F6], the module is ${\mathrm{\Omega }}_{L\mathrm{/}k}=L\mathrm{/}(0)=L$, generated by $\mathrm{d}t$; explicitly ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\cdot \mathrm{d}t$ is one-dimensional over the field $L$ and nonzero. Since $t^p=a$ with $a\in k\setminus k^p$, the element $t$ is not separable over $k$, so this is a field extension with a nonzero differential module, in contrast with clause 1.

The chart on $D(y)$. Take $R=k$, $n=2$, $c=1$, the ordered variables $(x_1,x_2)=(y,x)$, the equation $f_1=y^2-x^3$ and $g=y$; then $S=A_y\cong (k[x_1,x_2]\mathrm{/}(f_1){)}_g$ by construction [F1]. By [F2] the Jacobian row of the single equation is $({\mathrm{\partial }}_{y}f,{\mathrm{\partial }}_{x}f)=(2y,-3x^2)$, and its first entry is $2y$, a product of the unit $2\in k$ and the unit $y$ of $S=A_y$; hence the leading $1\times 1$ minor is a unit of $S$ and the presentation is standard smooth of relative dimension $2-1=1$. This proves the claim on the whole open set $D(y)=\{y\mathrm{\ne }0\}\subseteq \mathrm{Spec}A$.

Complements. The hypothesis on the characteristic is exactly what the unit computation uses: if $2=0$ in $k$ then $2y=0$ is not a unit of $S$. At the origin both partial derivatives $2y$ and $-3x^2$ vanish, so the displayed single equation gives no invertible minor there, and the chart of step 1.1 covers precisely the points with $y\mathrm{\ne }0$, not the cusp at the origin.

The ring $L$ is a field. Choose a monic irreducible factor $Q$ of $T^p-a$ by [F1] and let $\xi$ be the class of $T$ in the field $F=k[T]\mathrm{/}(Q)$. Then ${\xi }^p=a$, and [F2] gives $T^p-a=(T-\xi {)}^p$ in $F[T]$. Thus $Q$ has only one distinct root in a splitting field. If $\mathrm{deg}Q=1$, then $\xi \in k$ contradicts $a\mathrm{\notin }{k}^p$. If $Q^{\mathrm{\prime }}\mathrm{\ne }0$, irreducibility and [F1] would make $Q$ separable with $\mathrm{deg}Q\ge 2$ distinct roots, also impossible. Thus $Q^{\mathrm{\prime }}=0$, so all exponents of $Q$ are divisible by $p$. Since $1\le \mathrm{deg}Q\le p$ and $Q$ is monic, it has degree $p$; as a monic divisor of $T^p-a$ of that degree it equals $T^p-a$. Hence $L$ is a field by [F1].

The tensor product. By [F3] there is an isomorphism $L{\otimes }_{k}L\cong L[T]\mathrm{/}(T^p-a)$, the second factor acting on coefficients; by [F2] one has $T^p-a=(T-\alpha {)}^p$ in $L[T]$, so substituting $u=T-\alpha$, an automorphism of $L[T]$, gives $L{\otimes }_{k}L\cong L[u]\mathrm{/}(u^p),$ under which $u$ corresponds to $T\otimes 1-1\otimes \alpha$.

The element $u$ is a nonzero nilpotent. In $L[u]\mathrm{/}(u^p)$ the classes of $1,u,\dots ,u^{p-1}$ are an $L$-basis, because $u^p$ is monic of degree $p$ and division with remainder is available: hence $u\mathrm{\ne }0$ while $u^p=0$. Consequently $L{\otimes }_{k}L$ is not reduced, so the base change of the field $L$ along $k\to L$ is a nonreduced local ring with residue field $L$, and the fibre is a thickening rather than a reduced point.

The element $\alpha$ and the nilpotent $u$. Since $L\mathrm{\ne }k$, choose $\alpha \in L\setminus k$; by [F1] there is $n\ge 1$ with $a:={\alpha }^{p^n}\in k$, and we fix such an $n$ (the minimal one). Put $u:=\alpha \otimes 1-1\otimes \alpha \in L{\otimes }_{k}L$.

$u\mathrm{\ne }0$. Since $L$ is finite-dimensional over $k$ and $\alpha \mathrm{\notin }k$, the elements $1,\alpha$ are $k$-linearly independent, so the linear functional on the plane $k\cdot 1+k\cdot \alpha$ with $\lambda (1)=1$, $\lambda (\alpha )=0$ extends to a $k$-linear map $\lambda \text{\hspace{0.17em}⁣}:L\to k$ (finite-dimensional linear algebra). If $u=0$, then applying $\lambda \otimes \mathrm{id}$ to the identity $\alpha \otimes 1=1\otimes \alpha$ gives $\lambda (\alpha )\cdot 1=\lambda (1)\cdot \alpha$, that is $0=\alpha$, a contradiction; hence $u\mathrm{\ne }0$.

$u$ is nilpotent. By [F2], in the commutative ring $L{\otimes }_{k}L$ of characteristic $p$ one has $u^{p^n}=(\alpha \otimes 1{)}^{p^n}-(1\otimes \alpha {)}^{p^n}={\alpha }^{p^n}\otimes 1-1\otimes {\alpha }^{p^n}$, and this is $a\otimes 1-1\otimes a=0$ because $a\in k$ satisfies $a\otimes 1=1\otimes a$ by [F5]. So $u\mathrm{\ne }0$ is a nilpotent element and $L{\otimes }_{k}L$ is not reduced.

$L$ is regular and $L{\otimes }_{k}L$ is not. The field $L$ is Noetherian and its only prime is $(0)$, whose localisation is the field $L$ itself: a field is a regular local ring of dimension $0$ with zero maximal ideal, so $\mathrm{edim}=\dim =0$ by [F3], and $L$ is regular. Suppose $L{\otimes }_{k}L$ were regular. It is a nonzero finite-dimensional $k$-algebra, so some maximal ideal $\mathfrak{m}$ contains the annihilator of $u$ and then $u$ has nonzero image in the localisation at $\mathfrak{m}$; that localisation would be a regular local ring, hence a domain by [F4], in which the nilpotent image of $u$ must vanish, a contradiction. Therefore $L{\otimes }_{k}L$ is not regular, so regularity of the two field factors $L$ and $L$ does not pass to their tensor product over $k$.

The standard smooth chart. The source coordinate ring is the polynomial ring $k[y_1,\dots ,y_m][x_1,\dots ,x_n]$, and the map $k[y_1,\dots ,y_m]\to k[y][x]$ is the structure map of the target algebra over itself, presented by $n$ variables, no equations ($c=0$) and $g=1$; by [F1] this is a standard smooth presentation of relative dimension $n-0=n$, so $\pi$ is standard smooth in this single chart, with no minor to check.

The cotangent parameters. At the $k$-rational point $(a,b)$ the residue field is $k$, so [F3] identifies the cotangent space with ${\mathrm{\Omega }}_{k[y,x]\mathrm{/}k}\otimes k$, which by [F2] has the $k$-basis $\mathrm{d}{y}_1,\dots ,\mathrm{d}{y}_m,\mathrm{d}{x}_1,\dots ,\mathrm{d}{x}_n$ and hence the classes of the coordinate differences $y_i-a_i$, $x_j-b_j$ as its $k$-basis. The pullback map on cotangent spaces induced by $\pi$ sends the class of $y_i-a_i$ to the class of ${\pi }^{\mathrm{\#}}(y_i-a_i)=y_i-a_i$, that is, it carries the basis $\mathrm{d}{y}_1,\dots ,\mathrm{d}{y}_m$ of the target cotangent space onto the first $m$ elements of a basis of the source cotangent space; in particular these classes are $k$-linearly independent.

The fibres. Let $a\in {\mathbf{A}}_k^m(k)$ be a $k$-rational point and let $I=(y_1-a_1,\dots ,y_m-a_m)\subseteq k[y]$ be its maximal ideal, with $k[y]\mathrm{/}I\cong k$. By [F4] the fibre ring is $k[y,x]{\otimes }_{k[y]}k\cong k[y,x]\mathrm{/}Ik[y,x]\cong k[x_1,\dots ,x_n]$, so the fibre over $a$ is $\mathrm{Spec}k[x_1,\dots ,x_n]={\mathbf{A}}_k^n$ and has dimension $n$ by [F5]; equivalently the fibre of $\pi$ over any $k$-rational point is affine $n$-space. This completes the verification of all three clauses, and the fibre dimension $n$ is exactly the relative dimension of the chart of step 1.1, while the target's $m$ coordinate parameters pull back to independent cotangent classes by step 1.2.